A particle moves in the XY plane under the . . .

Question : A particle moves in the XY plane under the influence of a force such that its linear momentum is p(t) = A[icos(kt)-jsin(kt)] where A and k are constants. The angle between the force and momentum vector is

a) 90°

b) 45°

c) 30°

d) 0°

Doubt by Muskan

Solution : 

p(t) = A[icos(kt)-jsin(kt)] — (1)

We know 

F = dp/dt

F = d{A[icos(kt)-jsin(kt)]}/dt

F = A[-iksin(kt)-jksin(kt)]

F = Ak[-isin(kt)-jsin(kt)] — (2)

Now, multiplying eq (1) and eq (2) by dot product. 

F.p 

= {Ak[-isin(kt)-jsin(kt)]}.{A[icos(kt)-jsin(kt)]}

= kA²[-sin(kt) cos(kt) + sin(kt) cos(kt)]

= kA²[0]
=0

∵F.p = 0

It means F丄 p
so, angle between F and p is 90°.

Hence, a) would be the correct option.